The function, f ( x ) = e x at 11 equidistant points on the interval [ 0 , 1 ].The question asks whether the trapezoidal rule on 10 subintervals would give a better approximation than the Simpson rule on 10 subintervals on the provided function?

Question

The function,  f  (  x  )  =      e    x  at   11 equidistant points on the interval   [  0  ,  1  ].The question asks whether the trapezoidal rule on   10 subintervals would give a better approximation than the Simpson rule on   10 subintervals on the provided function?

Duncan Cox · Accepted Answer

For Simpson&#039;s rule, S(f), there exists a point ξ∈[0,1] such that the error  E  (  S  (  f  )  ,  n  )  =      |          ∫      0      1        f    (    x    )    d    x    −    S    (    f    )    |    =      1          180              n        4                  |        f          (      4      )        (  ξ  )      |    ≤      e          180              n        4              .For the trapezoidal rule,   T  (  f  ), the error is  E  (  T  (  f  )  ,  n  )  =      |          ∫      0      1        f    (    x    )    d    x    −    T    (    f    )    |    =      1          12              n        2                  |        f          &quot;        (  ξ  )      |    ≥      1          12              n        2              .We thus have  E  (  S  (      e    x    )  ,  5  )  ≤  3  e  −  05and  E  (  T  (      e    x    )  ,  10  )  ≥  8  e  −  04  ,hence we can be sure that Simpson&#039;s rule will be the better one. In fact, I got   E  (  S  (      e    x    )  ,  5  )  =  7.3415  e  −  06 and   E  (  T  (      e    x    )  ,  10  )  =  0.0012, so the difference is somewhat larger than the theoretical bounds predict.

The function, f ( x ) = e x </msup> at 11 equidistant poi

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